Problem: Find the limit as $x$ approaches negative infinity. $\lim_{x\to-\infty}\dfrac{3x}{\sqrt{16x^2-9x}}=$
Explanation: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the numerator is $x$, let's divide by $x$. In the denominator, let's divide by $-\sqrt{x^2}$, since for negative values, $x=-\sqrt{x^2}$. $\begin{aligned} &\phantom{=}\lim_{x\to-\infty}\dfrac{3x}{\sqrt{16x^2-9x}} \\\\ &=\lim_{x\to-\infty}\dfrac{\dfrac{3x}{x}}{\dfrac{\sqrt{16x^2-9x}}{-\sqrt{x^2}}} \gray{\text{Divide sides by }x=-\sqrt{x^2}} \\\\ &=\lim_{x\to-\infty}-\dfrac{\dfrac{3x}{x}}{\dfrac{\sqrt{16x^2-9x}}{\sqrt{x^2}}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to-\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to-\infty}-\dfrac{\dfrac{3\cancel{x}}{\cancel{x}}}{\sqrt{\dfrac{16\cancel{x^2}}{\cancel{x^2}}-\dfrac{9\cancel x}{\cancel x \cdot x}}} \\\\ &=\lim_{x\to-\infty}-\dfrac{3}{\sqrt{16-\dfrac{9}{x}}} \\\\ &=\lim_{x\to-\infty}-\dfrac{3}{\sqrt{16-0}} \gray{\lim_{x\to-\infty}\dfrac{k}{x^n}=0} \\\\ &=-\dfrac{3}{\sqrt{16}} \\\\ &=-\dfrac{3}{4} \end{aligned}$ In conclusion, $\lim_{x\to-\infty}\dfrac{3x}{\sqrt{16x^2-9x}}=-\dfrac{3}{4}$.